snprintf for string concatenation

Question

I am using snprintf to concatenate a string to a char array:

char buf[20] = "";
snprintf(buf, sizeof buf, "%s%s", buf, "foo");
printf("%s\n", buf);
snprintf(buf, sizeof buf, "%s%s", buf, " bar");
printf("%s\n", buf);

The problem is the second concatenation to buf instead of adding "bar", replaces "foo" with it. The output is like:

foo
bar

The first %s should keep buf (which in this case holds "foo") there. And the second %s should attach "bar" to it. Right?

What am I doing wrong?

Answer 1

You're violating the restrict contract on snprintf, which states that no other argument can overlap the buffer.

Copying the input into itself is a waste of effort anyway. snprintf returns the number of characters which formatting would require, so take advantage of this for appending:

char buf[20] = "";
char *cur = buf, * const end = buf + sizeof buf;
cur += snprintf(cur, end-cur, "%s", "foo");
printf("%s\n", buf);
if (cur < end) {
    cur += snprintf(cur, end-cur, "%s", " bar");
}
printf("%s\n", buf);

Answer 2

While the accepted answer is alright, the better (in my opinion) answer is that concatenating strings is wrong. You should construct the entire output in a single call to snprintf. That's the whole point of using formatted output functions, and it's a lot more efficient and safer than doing pointer arithmetic and multiple calls. For example:

snprintf(buf, sizeof buf, "%s%s%s", str_a, str_b, str_c);